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In the diagram above, angle C = 90º and AC = BC. Point M is the midpo

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In the diagram above, angle C = 90º and AC = BC. Point M is the midpo  [#permalink]

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New post 11 Mar 2015, 04:56
3
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A
B
C
D
E

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  75% (hard)

Question Stats:

69% (03:28) correct 31% (03:33) wrong based on 172 sessions

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In the diagram above, angle C = 90º and AC = BC. Point M is the midpoint of AB. Arc AXB has its center at C, and passes through A and B. Arc AYB has its center at M and passes through A and B. The shaded region between the two arcs is called a “lune.” What is the ratio of the area of the lune to the area of triangle ABC?

A. 1
B. \(\frac{\pi}{2}\)
C. \(\frac{\pi}{3}\)
D. \(\frac{3\pi}{8}\)
E. \(\frac{4\pi}{9}\)

Kudos for a correct solution.

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Re: In the diagram above, angle C = 90º and AC = BC. Point M is the midpo  [#permalink]

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New post 11 Mar 2015, 05:58
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Hi
A very time consuming
have edited the pic for all the forum members
Answer in the pic attached

cheers..!!
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Re: In the diagram above, angle C = 90º and AC = BC. Point M is the midpo  [#permalink]

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New post 15 Mar 2015, 21:23
Bunuel wrote:
Image
In the diagram above, angle C = 90º and AC = BC. Point M is the midpoint of AB. Arc AXB has its center at C, and passes through A and B. Arc AYB has its center at M and passes through A and B. The shaded region between the two arcs is called a “lune.” What is the ratio of the area of the lune to the area of triangle ABC?

A. 1
B. \(\frac{\pi}{2}\)
C. \(\frac{\pi}{3}\)
D. \(\frac{3\pi}{8}\)
E. \(\frac{4\pi}{9}\)

Kudos for a correct solution.


MAGOOSH OFFICIAL SOLUTION:

This problem is based on a famous theorem of Hippocrates of Chios (c. 470 – c. 410 BCE), a predecessor of Euclid. The easiest way to see this is as follows.

Let AC = BC = 2S. This is the radius of the larger circle, which has an area of 4pi*S^2, so that the quarter circle:
Image
must have an area of pi*S^2. Hold that thought.

Now, notice that triangle ABC is a 45-45-90 triangle, so that the length of the hypotenuse must be
\(AB=2\sqrt{2}S\)
and the radius of the smaller circle is
\(AM=\sqrt{2}S\)

The area of that whole circle would be 2pi*S^2, and the area of the semicircle:
Image

would be half of that, pi*S^2. Thus, the quarter circle of the larger circle and semicircle of the smaller circle have the same area. That’s a very big idea!

Now, look at the circular segment:
Image


We don’t need to know that area: simply call it J.

If we subtract the circular segment, J, from the quarter circle of the bigger circle, we get the triangle:
pi*S^2 – J = area of triangle ABC

If we subtract the circular segment, J, from the semicircle of the smaller circle, we get the lune:
pi*S^2 – J = area of the lune

Because those two areas equal the same thing, they must equal each other.
area of triangle ABC = area of the lune

The two are equal, so their ratio is 1.

Answer = (A)
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Re: In the diagram above, angle C = 90º and AC = BC. Point M is the midpo  [#permalink]

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New post 21 Nov 2016, 20:03
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Let set \(AC=BC=x\)

We have \(S_{ABC}=\frac{x^2}{2}\)

\(AXBC\) is a quarter circle, so \(S_{AXBC}=\frac{\pi x^2}{4}\)

Hence \(S_{AMBX}=S_{AXBC}-S_{ABC}=\frac{\pi x^2}{4}-\frac{x^2}{2}\)

We have \(AB=x \sqrt{2}\) so \(AM=MB=\frac{x}{\sqrt{2}}\)

\(AYBM\) is a semi-circle, so \(S_{AYBM}=\frac{1}{2} \times \pi \times ( \frac{x}{\sqrt{2}} )^2=\frac{\pi x^2}{4}\)

Hence \(S_{AYBX}=S_{AYBM}-S_{AMBX}=\frac{\pi x^2}{4} - (\frac{\pi x^2}{4}-\frac{x^2}{2})=\frac{x^2}{2}\)

This means \(S_{AYBX}=S_{ABC}\).

The answer is A
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Re: In the diagram above, angle C = 90º and AC = BC. Point M is the midpo  [#permalink]

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New post 22 Aug 2018, 11:07
Don't mess with algebra. Assume that AC=BC=1. Calculations will be much easier and faster.
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Re: In the diagram above, angle C = 90º and AC = BC. Point M is the midpo   [#permalink] 22 Aug 2018, 11:07
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